Chemin J.-Y., Desjardins B., Gallagher I., Grenier E. Mathematical geophysics
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Non-linear estimates in the well-prepared case 169
recalling that v
3
0,int
=
v
3
N
= 0. But
F
ε,η
1
L
2
([0,T ];L
2
)
≤v
ε,h
app
L
∞
([0,T ];L
∞
)
∇
h
(v
ε
app
− v
N
)
L
2
([0,T ];L
2
)
.
Using the estimate (7.1.25), we infer that
lim sup
ε→0
F
ε,η
1
L
2
([0,T ];L
2
)
≤ C lim sup
ε→0
∇
h
(v
ε
app
−
v
N
)
L
2
([0,T ];L
2
)
,
and it is just a matter of using (7.1.28) to conclude. The result on F
ε,η
2
is proved
along the same lines.
In order to estimate F
ε,η
3
, let us write
F
ε,η
3
L
2
([0,T ];L
2
)
≤v
ε,3
app
L
∞
([0,T ];L
∞
)
∂
3
v
ε
app
L
2
([0,T ];L
2
)
.
By definition (7.1.11), we have
v
ε,3
app
L
∞
([0,T ];L
∞
)
≤ C
η
ε.
We infer that
F
ε,η
3
L
2
([0,T ];L
2
)
≤ C
η
ε∂
3
v
ε
app
L
2
([0,T ];L
2
)
.
The energy estimate (7.1.25) implies that ε
1
2
∂
3
v
ε
app
L
2
([0,T ];L
2
)
is uniformly
bounded, which implies the expected result on F
ε,η
3
and the lemma is proved.
Now let us prove the following result.
Lemma 7.3 Let
v be a solution of (SE
ν,β
) and let η be a posit-
ive real number. Denote by (v
ε,η
app
)
ε>0
the families given by Lemma 7.1.
Then for any vector field δ belonging to L
∞
([0,T]; H) ∩ L
2
([0,T]; V
σ
),
we have
T
0
δ(t) ·∇δ(t)|v
ε,η
app
(t)
L
2
dt ≤
C
η
ε
1
2
+
1
4
E
ε
T
(δ)
+
C
ν
T
0
∇
h
v(t)
2
L
2
δ(t)
2
L
2
dt.
Proof The proof of the result relies on the explicit expression of (v
ε,η
app
)
ε>0
,
jointly with the following lemma.
Lemma 7.4 Let δ be a vector field in V
σ
and let w be a bounded vector field.
Then
(δ ·∇δ|w)
L
2
≤ C∇
h
δ
L
2
∂
3
δ
L
2
d
1
2
(·)w
L
2
(]0,1[;L
∞
h
)
,
where d denotes the distance to the boundary.
170 Ekman boundary layers for rotating fluids
Proof Let us define
I
j,k
=
Ω
δ
k
∂
k
δ
j
w
j
dx,
for j and k in {1, 2, 3}. In the case k = 3, since δ vanishes at the boundary we
can write that
|δ
k
(x
h
,x
3
)| =
x
3
0
∂
3
δ
k
(x
h
,y
3
) dy
3
≤ x
1
2
3
∂
3
δ
L
2
(]0,1[)
,
and similarly for the upper boundary x
3
= 1. We therefore have
|δ
k
(x
h
,x
3
)|≤d(x
3
)
1
2
∂
3
δ(x
h
, ·)
L
2
(]0,1[)
.
We infer that for k =3,
|I
j,k
|≤
Ω
∂
3
δ(x
h
, ·)
L
2
(]0,1[)
|∇
h
δ(x)| d(x
3
)
1
2
|w(x)| dx
≤
Ω
∂
3
δ(x
h
, ·)
L
2
(]0,1[)
|∇
h
δ(x)| d(x
3
)
1
2
w(·,x
3
)
L
∞
h
dx.
The Cauchy–Schwarz inequality implies finally that for k =3,
|I
j,k
|≤∂
3
δ
L
2
∇
h
δ
L
2
d
1
2
(·)w
L
2
(]0,1[;L
∞
h
)
.
Now let us turn to the case k = 3. Using the fact that δ is divergence-free, we get
δ
3
(x
h
,x
3
)=
x
3
0
∂
3
δ
3
(x
h
,y
3
) dy
3
= −
x
3
0
div
h
δ
h
(x
h
,y
3
) dy
3
.
It follows as above that
|δ
3
(x
h
,x
3
)|≤d(x
3
)
1
2
∇
h
δ(x
h
, ·)
L
2
(]0,1[)
,
which implies that
|I
j,3
|≤
Ω
∇
h
δ(x
h
, ·)
L
2
(]0,1[)
|∂
3
δ(x)| d(x
3
)
1
2
|w(x)| dx.
The estimate is now the same as in the previous case k = 3. Thus the lemma is
proved.
Now let us go back to the proof of Lemma 7.3. We recall that by the definition
of v
ε,η
app
,wehave
v
ε,η
app
−
v
N
η
− v
0,BL
L
∞
([0,T ];L
∞
(Ω))
≤ C
η
ε.
Non-linear estimates in the well-prepared case 171
It follows that
T
0
δ(t) ·∇δ(t)|(v
ε,η
app
−
v
N
η
− v
0,BL
)(t)
dt ≤ C
η
ε
T
0
δ(t)
L
2
∇δ(t)
L
2
dt
which, using the fact that ε
1
2
∂
3
δ
L
2
([0,T ]×Ω)
≤ E
ε
T
(δ), yields
T
0
δ(t) ·∇δ(t)|(v
ε,η
app
−
v
N
η
− v
0,BL
)(t)
dt ≤ C
η
ε
1
2
E
ε
T
(δ). (7.2.1)
Moreover
|v
0,BL
(x)|≤C|
v
N
η
(x
h
)|
exp
−
x
3
ε
√
2β
+exp
−
1 − x
3
ε
√
2β
.
Thus
d(x
3
)
1
2
v
0,BL
(·,x
3
)
L
∞
h
≤ CN
η
v
N
η
L
2
x
1
2
3
exp
−
x
3
ε
√
2β
+(1−x
3
)
1
2
exp
−
1 − x
3
ε
√
2β
.
A classical L
2
energy estimate on (SE
ν,β
) implies that
v
N
η
2
L
∞
(R
+
;L
2
)
≤
v
0
2
L
2
+
2
ν
f
2
L
2
([0,T ];H
−1,0
)
,
so we deduce that
d(x
3
)
1
2
v
0,BL
2
L
2
(]0,1[;L
∞
)
≤ 2C
0
N
2
η
ε
2
x
3
ε
exp
−2
x
3
ε
√
2β
dx
3
ε
,
where C
0
denotes any O
v
0
2
L
2
+
2
ν
f
2
L
2
([0,T ];H
−1,0
)
. Finally
d(x
3
)
1
2
v
0,BL
2
L
2
(]0,1[;L
∞
)
≤ C
0
N
2
η
ε
2
. (7.2.2)
Now let us write that
T
0
(δ(t) ·∇δ(t)|v
ε,η
app
(t))
L
2
dt ≤
3
j=1
V
j
(T ) where
V
1
(T )
def
=
T
0
δ(t) ·∇δ(t)|v
ε,η
app
−
v
N
η
− v
0,BL
L
2
dt ,
V
2
(T )
def
=
T
0
(δ(t) ·∇δ(t)|v
0,BL
(t))
L
2
dt and
V
3
(T )
def
=
T
0
δ(t) ·∇δ(t)|
v
N
η
(t)
L
2
dt.
172 Ekman boundary layers for rotating fluids
Estimate (7.2.1) claims exactly that
V
1
(T ) ≤ C
η
ε
1
2
E
ε
T
(δ).
Lemma 7.4, together with (7.2.2), imply that
V
2
(T ) ≤ C
η
ε
T
0
∇
h
δ(t)
L
2
∂
3
δ(t)
L
2
d(x
3
)
1
2
v
0,BL
(t)
L
2
(]0,1[;L
∞
)
dt
≤ C
η
∇
h
δ
L
2
([0,T ];L
2
)
ε
1
2
∂
3
δ
L
2
([0,T ];L
2
)
.
By definition of E
ε
t
we infer, using the Cauchy–Schwarz inequality, that
V
2
(T ) ≤ C
η
ε
1
2
E
ε
T
(δ).
Thus we get that
T
0
(δ(t) ·∇δ(t)|v
ε,η
app
(t))
L
2
dt ≤V
3
(T )+C
η
ε
1
2
E
ε
T
(δ).
Now we are left with the estimate of
V
3
(T )=
T
0
|(δ(t) ·∇δ(t)|
v
N
η
(t))
L
2
|dt.
As δ is divergence-free and vanishes at the boundary, we have
(δ ·∇δ|
v
N
η
)
L
2
= −(δ
h
·∇
h
v
N
η
|δ)
L
2
≤∇
h
v
N
η
L
2
1
0
δ(·,x
3
)
2
L
4
h
dx
3
.
The two-dimensional Gagliardo–Nirenberg and Cauchy–Schwarz inequalities
imply that
V
3
(T ) ≤ C
T
0
∇
h
v
N
η
(t)
L
2
δ(t)
L
2
∇
h
δ(t)
L
2
dt
≤
ν
2
T
0
∇
h
δ(t)
2
L
2
dt +
C
ν
T
0
∇
h
v
N
η
(t)
2
L
2
δ(t)
2
L
2
dt (7.2.3)
which ends the proof of Lemma 7.3.
7.3 The convergence theorem in the well-prepared case
In this section we intend to prove Theorem 7.1. The idea is to apply Lemma 7.1
to
v
def
= u. Indeed u solves a system of the type (SE
ν,β
), stating f
def
= Q(u, u) with
the notation of Definition 2.7, page 44. We recall that according to Lemma 2.3,
Q(
u, u)isanelementofL
2
(R
+
; V
′
σ
) as long as u is in the energy space.
Let T>0 be given and let η be an arbitrarily small positive real number.
According to Lemma 7.1, the theorem will be proved if we prove that
sup
t∈[0,T ]
(u
ε
− u
ε,η
app
)(t)
2
L
2
(Ω)
+ ν
T
0
∇
h
(u
ε
− u
ε,η
app
)(t)
2
L
2
(Ω)
dt = ρ
ε,η
,
The convergence theorem in the well-prepared case 173
where u
ε,η
app
is the family given by Lemma 7.1, associated with η,
v = u and f = Q(u, u)=−u ·∇
h
u.
The main step of the proof is purely algebraic and is common to this case and
to the ill-prepared case (in the R
2
case in Section 7.6 as well as in the periodic
case in Section 7.7). It is based on the following lemma.
Lemma 7.5 Let (u
ε
) be a family of Leray solutions of (NSC
ε
) with initial
data u
0
, and let (Ψ
ε
)
ε>0
be a family of C
1
(R
+
; V
σ
(Ω)) functions. Then the
function δ
ε
def
= u
ε
− Ψ
ε
satisfies
E
ε
t
(δ
ε
)=E
ε
t
(u
ε
)+E
ε
t
(Ψ
ε
) − 2(u(0)|Ψ
ε
(0))
L
2
(7.3.1)
−2
t
0
(δ
ε
(t
′
) ·∇δ
ε
(t
′
)|Ψ
ε
(t
′
))
L
2
dt
′
+2
t
0
(G
ε
(Ψ
ε
)(t
′
)|u
ε
(t
′
))
L
2
dt
′
,
where G
ε
(Ψ
ε
)
def
= L
ε
Ψ
ε
+Ψ
ε
·∇Ψ
ε
.
Before proving the lemma, let us show why it leads to Theorem 7.1. We apply
Lemma 7.5 with Ψ
ε
= u
ε,η
app
and u
0
= u
0
. Note that preliminary smoothing in
time is required, so as to have u
ε,η
app
∈ C
1
(R
+
, V
σ
). That procedure should by now
be familiar to the reader, and in order to avoid introducing additional notation
we shall omit that time smoothing procedure in the following. As u
ε
is a Leray
solution of (NSC
ε
), it satisfies
E
ε
t
(u
ε
) ≤
u
0
2
L
2
.
Furthermore, estimate (7.1.25) yields
E
ε
t
(u
ε,η
app
) ≤
u
0
2
L
2
+2
t
0
u(t
′
) ·∇
h
u(t
′
), u(t
′
)dt
′
+ ρ
ε,η
= u
0
2
L
2
+ ρ
ε,η
and since
u
ε,η
app|t=0
−
u
0
L
2
= ρ
ε,η
,
we get
E
ε
t
(u
ε
)+E
ε
t
(u
ε,η
app
) − 2(u
ε
(0)|u
ε,η
app|t=0
)
L
2
≤ ρ
ε,η
. (7.3.2)
Then Lemma 7.1 implies that
L
ε
u
ε,η
app
= −
u ·∇
h
u + R
ε,η
+ ∇p
ε
,
where R
ε,η
denotes from now on generically any vector field satisfying
R
ε,η
L
2
([0,T ];H
−1,0
)
= ρ
ε,η
.
It follows that one can write
G
ε,η
(u
ε,η
app
)=u
ε,η
app
·∇u
ε,η
app
−
u
N
η
·∇
h
u
N
η
+
u
N
η
·∇
h
u
N
η
−
u ·∇
h
u + R
ε,η
+ ∇p
ε
.
174 Ekman boundary layers for rotating fluids
As Q is a continuous bilinear map from
L
4
([0,T]; L
4
h
) × L
4
([0,T]; L
4
h
)intoL
2
([0,T]; H
−1
(Ω
h
))
we get
G
ε,η
(u
ε,η
app
)=u
ε,η
app
·∇u
ε,η
app
−
u
N
η
·∇
h
u
N
η
+ R
ε,η
+ ∇p
ε
.
By Lemma 7.2 we find
G
ε,η
(u
ε,η
app
)=F
ε,η
+ R
ε,η
+ ∇p
ε
(7.3.3)
with lim
ε→0
F
ε,η
L
2
([0,T ],L
2
(Ω))
= 0. Since
t
0
(R
ε,η
(t
′
)|u
ε
(t
′
))
L
2
dt
′
≤R
ε,η
L
2
([0,T ],H
−1,0
)
∇
h
u
ε
L
2
([0,T ],L
2
)
= ρ
ε,η
,
(7.3.4)
and
t
0
(F
ε,η
(t
′
)|u
ε
(t
′
))
L
2
dt
′
≤ t
1
2
u
0
L
2
F
ε,η
L
2
([0,T ];L
2
)
= ρ
ε,η
, (7.3.5)
plugging both inequalities into (7.3.3) yields
T
0
(G
ε,η
(u
ε,η
app
)(t
′
)|u
ε
(t
′
))
L
2
dt
′
≤ ρ
ε,η
. (7.3.6)
Putting (7.3.2) and (7.3.6) into (7.3.1) yields finally
E
ε
t
(δ
ε
) ≤ C
t
0
δ
ε
(t
′
) ·∇δ
ε
(t
′
)|u
ε,η
app
(t
′
)
L
2
dt
′
+ ρ
ε,η
.
It is now just a matter of using Lemma 7.3 to obtain, for ε small enough compared
to N , that
E
ε
t
(δ
ε
) ≤ ρ
ε,η
+
C
ν
t
0
∇
h
v(t
′
)
2
L
2
δ
ε
(t
′
)
2
L
2
dt
′
,
and the result follows from Gronwall’s inequality: we find
E
ε
t
(δ
ε
) ≤ ρ
ε,η
exp
C
0
ν
2
= ρ
ε,η
and Theorem 7.1 is proved.
Proof of Lemma 7.5 This is a typical weak–strong type argument along the
lines of those followed in Chapter 3 in the proof of the stability Theorem 3.3 (see
pages 59–63). We have
E
ε
t
(δ
ε
)=E
ε
t
(u
ε
)+E
ε
t
(Ψ
ε
) − 2B
ε
t
(u
ε
, Ψ
ε
), (7.3.7)
The convergence theorem in the well-prepared case 175
where
B
ε
t
(a, b)
def
=(a(t)|b(t))
L
2
+2ν
t
0
∇
h
a(t
′
)|∇
h
b(t
′
)
L
2
dt
′
+2εβ
t
0
(∂
3
a(t
′
)|∂
3
b(t
′
))
L
2
dt
′
.
Let us compute B
ε
t
(u
ε
, Ψ
ε
). Using Ψ
ε
as a test function in Definition 2.5, page 42,
we get
(u
ε
(t)|Ψ
ε
(t))
L
2
=(u
ε
(0)|Ψ
ε
(0))
L
2
− ν
t
0
(∇
h
u
ε
(t
′
)|∇
h
Ψ
ε
(t
′
))
L
2
dt
′
− εβ
t
0
(∂
3
u
ε
(t
′
)|∂
3
Ψ
ε
(t
′
))
L
2
dt
′
−
t
0
(u
ε
(t
′
) ·∇u
ε
(t
′
)|Ψ
ε
(t
′
))
L
2
dt
′
+
t
0
(u
ε
(t
′
)|∂
t
Ψ
ε
(t
′
))
L
2
dt
′
−
1
ε
t
0
(Ru
ε
(t
′
)|Ψ
ε
(t
′
))
L
2
dt
′
.
By definition of L
ε
Ψ
ε
,wehave
∂
t
Ψ
ε
= L
ε
Ψ
ε
+
ν∆
h
+ εβ∂
2
3
−
1
ε
R
Ψ
ε
,
hence by an integration by parts allowed by the fact that u
ε
and Ψ
ε
vanish at
the boundary, we infer that
(u
ε
|∂
t
Ψ
ε
)
L
2
= − ν(∇
h
u
ε
|∇
h
Ψ
ε
)
L
2
− εβ(∂
3
u
ε
|∂
3
Ψ
ε
)
L
2
−
1
ε
(u
ε
|RΨ
ε
)
L
2
+(u
ε
|L
ε
Ψ
ε
)
L
2
.
It follows that
(u
ε
(t)|Ψ
ε
(t))
L
2
=(u
ε
(0)|Ψ
ε
(0))
L
2
− 2ν
t
0
(∇
h
u
ε
(t
′
)|∇
h
Ψ
ε
(t
′
))
L
2
dt
′
− 2εβ
t
0
(∂
3
u
ε
(t
′
)|∂
3
Ψ
ε
(t
′
))
L
2
dt
′
−
t
0
(u
ε
·∇u
ε
(t
′
)|Ψ
ε
(t
′
))
L
2
dt
′
−
1
ε
t
0
(Ru
ε
(t
′
)|Ψ
ε
(t
′
))
L
2
dt
′
−
1
ε
t
0
(RΨ
ε
(t
′
)|u
ε
(t
′
))
L
2
dt
′
+
t
0
(u
ε
(t
′
)|L
ε
Ψ
ε
(t
′
))
L
2
dt
′
176 Ekman boundary layers for rotating fluids
so
B
ε
t
(u
ε
, Ψ
ε
)=(u
ε
(0)|Ψ
ε
(0))
L
2
(7.3.8)
−
t
0
(u
ε
·∇u
ε
(t
′
)|Ψ
ε
(t
′
))
L
2
dt
′
+
t
0
(u
ε
(t
′
)|L
ε
Ψ
ε
(t
′
))
L
2
dt
′
.
But relation (3.2.2) page 57, implies that
−(u
ε
·∇u
ε
|Ψ
ε
)
L
2
=(Q(u
ε
,u
ε
)|Ψ
ε
− u
ε
)
L
2
=(Q(Ψ
ε
, Ψ
ε
)|δ
ε
)
L
2
+(Q(δ
ε
,u
ε
)|δ
ε
)
L
2
= −(Ψ
ε
·∇Ψ
ε
|δ
ε
)
L
2
+(δ
ε
·∇δ
ε
|u
ε
)
L
2
. (7.3.9)
We then just have to put together (7.3.8), (7.3.9), and (7.3.7) to prove the lemma.
7.4 The ill-prepared linear problem
The goal of this section is to construct approximate solutions to
(SC
ε
β
)
∂
t
v
ε
− ν∆
h
v
ε
− βε∂
2
3
v
ε
+
e
3
∧v
ε
ε
= −∇p
ε
div v
ε
=0
v
ε
|t=0
= v
0
v
ε
|∂Ω
=0
in the case when the initial data v
0
does depend on the vertical variable x
3
.We
suppose throughout this section that Ω
h
= R
2
or T
2
and in the periodic case we
also suppose that the horizontal mean of v
ε
vanishes. We introduce the following
notation: H
0
(T
2
×[0, 1]) denotes the space of functions in H(T
2
×[0, 1]) of van-
ishing horizontal mean. In the following, to simplify notation we will denote by ξ
h
the horizontal Fourier variable (which in the periodic case belongs to Z
2
\{0}).
Let us recall that
L
ε
v
def
=
∂
t
v
h
− ν∆
h
v
h
+
Rv
h
ε
− βε∂
2
3
v
h
∂
t
v
3
− ν∆
h
v
3
− βε∂
2
3
v
3
and that R denotes rotation by angle π/2 in the horizontal plane.
The vector field v
ε
, which of course also depends on the vertical variable x
3
,
does not belong to the kernel of P(e
3
∧v
ε
). Thus we have to deal with very fast
time oscillations.
As in the well-prepared case, we will need to truncate functions in frequency
space, hence to consider divergence vector fields which are a finite sum of trigo-
nometric functions and the horizontal Fourier transform of which is supported
in rings. Let us start by proving the following lemma.
The ill-prepared linear problem 177
Lemma 7.6 Let us consider the space B =
-
∞
N=1
B
N
of vector fields on Ω of
the form
v(x
h
,x
3
)=
v
0,h
(x
h
)
0
+
N
k
3
=1
v
k
3
,h
(x
h
) cos(k
3
πx
3
)
−
1
k
3
π
div
h
v
k
3
,h
(x
h
) sin(k
3
πx
3
)
where (v
k
3
,h
)
1≤k
3
≤N
are two-dimensional vector fields on R
2
(resp. on T
2
),
the Fourier transform of which has its support included in the ring C (1/N, N )
of R
2
, and such that div
h
v
0,h
=0. The space B is dense in H(R
2
×[0, 1])
(resp. H
0
(T
2
×[0, 1])).
Proof In order to prove this lemma, let us prove that the orthogonal space of B
in H(R
2
×[0, 1]) (resp. H
0
(T
2
×[0, 1])) is {0}. Let us consider a vector field w in
the orthogonal space of B. In particular, for any function ϕ in L
2
(Ω
h
) the Fourier
transform of which is supported in a ring C of R
2
(and of vanishing horizontal
mean if Ω
h
= T
2
), we have, for any k
3
∈ N \{0},
P
k
3
(ϕ)
def
=
w
(∇
h
ϕ cos(k
3
πx
3
), −
1
k
3
π
∆
h
ϕ sin(k
3
πx
3
))
L
2
=0.
By integration by parts and using the fact that w
3
vanishes in the boundary, we
get that
P
k
3
(ϕ)=−
Ω
div
h
w
h
(x
h
,x
3
)ϕ(x
h
) cos(k
3
πx
3
)dx
h
dx
3
−
1
(k
3
π)
2
Ω
∂
3
w
3
(x
h
,x
3
)∆
h
ϕ(x
h
) cos(k
3
πx
3
)dx
h
dx
3
.
Using the fact that w is divergence-free gives that
P
k
3
(ϕ)=
R
2
1
(k
3
π)
2
∆
h
ϕ − ϕ
(x
h
)
×
1
0
div
h
w
h
(x
h
,x
3
) cos(k
3
πx
3
)dx
3
dx
h
.
For any positive integer k
3
, the operator (k
3
π)
−2
∆
h
− Id is an isomorphism in
the space of functions belonging to L
2
(Ω
h
) and the Fourier transform of which
is included in the ring C
N
def
= {η ∈ R
2
/ |η|∈[N
−1
,N]}.Thusifw belongs
to the orthogonal space of B, for any given positive integer k
3
, we have for
any function ψ such that Supp
ψ ⊂C
N
and for any positive integer k
3
such
that k
3
≤ N,
Ω
h
ψ(x
h
)
1
0
div
h
w
h
(x
h
,x
3
) cos(k
3
πx
3
)dx
3
dx
h
=0.
When Ω
h
= R
2
, we note that the space of functions of L
2
(R
2
), the Fourier
transform of which is included in a ring, is a dense subspace of L
2
(R
2
). When Ω
h
178 Ekman boundary layers for rotating fluids
is equal to T
2
, that still holds for L
2
(T
2
) restricted to functions the horizontal
mean of which vanishes. So in both cases we get that, for any positive integer k
3
,
1
0
div
h
w
h
(x
h
,x
3
) cos(k
3
πx
3
)dx
3
=0.
As w
3
|∂Ω
= 0, we have, thanks to (7.0.1), that
1
0
div
h
w
h
(x
h
,x
3
)dx
3
=0.
Thus for any non-negative integer k
3
,weget
1
0
div
h
w
h
(x
h
,x
3
) cos(k
3
πx
3
)dx
3
=0
and thus div w
h
(x
h
,x
3
) = 0 on Ω. As the vector field w is divergence-free, we
have ∂
3
w
3
=0.Asw
3
(x
h
, 1) = w
3
(x
h
, 0)=0,wehavew
3
≡ 0onΩ.
But w is orthogonal to B; so for any function ϕ on R
2
, the Fourier transform
of which is included in the ring C
N
(and of vanishing horizontal mean if Ω
h
= T
2
),
we have for any non-negative integer k
3
w|(∇
h,⊥
ϕ cos(k
3
πx
3
), 0)
L
2
=0.
This means exactly that, for any non-negative integer k
3
,
R
2
ϕ(x
h
)
1
0
curl
h
w
h
(x
h
,x
3
) cos(k
3
πx
3
)dx
3
dx
h
.
Thus we have curl
h
w
h
= 0 on Ω. The lemma is proved.
Remarks • The choice of the basis (cos(k
3
πx
3
))
k
3
∈N
for the horizontal com-
ponent ensures that the boundary condition v
3
|∂Ω
= 0 is satisfied because
the vertical component is of the type sin(k
3
πx
3
) thanks to the divergence-free
condition.
• Let us note that for any v in
B,wehave
v
2
H
=
∞
k
3
=0
v
k
3
,h
2
L
2
(Ω
h
)
+
∞
k
3
=1
1
(k
3
π)
2
div
h
v
k
3
,h
2
L
2
(Ω
h
)
. (7.4.1)
This leads us naturally to the following definition. We define the sets
G =
w ∈ L
2
(Ω
h
; R
2
)
.
div
h
w ∈ L
2
(Ω
h
)
and
G
0
= G ∩
w ∈ L
2
(T
2
; R
2
)
.
T
2
w(x
h
) dx
h
=0
.